Calculating Force and Momentum: A Catcher's Challenge

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Homework Help Overview

The problem involves calculating the change in momentum and the force applied by a catcher to stop a baseball. The subject area includes concepts of momentum and force, particularly in the context of a collision scenario.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need for unit conversion from kilometers per hour to meters per second, highlighting the importance of correct units in calculations. There are attempts to clarify the application of impulse formulas and the relationships between force, mass, and velocity.

Discussion Status

Some participants have provided guidance on unit conversion and have pointed out potential mistakes in the original calculations. There is an ongoing exploration of the correct approach to calculating force and momentum, with some participants expressing confidence in revised calculations.

Contextual Notes

There is a noted emphasis on the challenges of unit conversion and the implications it has on the calculations. The original poster acknowledges mistakes and expresses a willingness to correct them.

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Homework Statement


A catcher stops a .250kg ball traveling at 100km/hr in .19s.
a. Calculate the change in momentum of the baseball [-6.9ns]
b. Calculate the force the catcher applies to the baseball [-36.5n]

Homework Equations


△p=m△v
f△t=m△v

The Attempt at a Solution


a. △p=m△v
△p=m(vi-vf)
△p=mvi-mvf
△p=(.25kg*100m/s)-(.25kg*0m/s)
△p=25ns

b. F△t=m△v
F=m△v-△t
F=((.25kg*100m/s)-(.25kg*0m/s))-(.19s)
F=24.81n

Both answers are wrong. Correct me please.
 
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In your attempt you have 100m/s but in your question, you have 100km/hr, you see the difference? Looks like you need to do a conversion.
 
You need to pay more attention to the units. Getting the units right in these problems is usually the most difficult part. 100 km/hr is not the same as 100 m/s.
 
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Likes   Reactions: RaulTheUCSCSlug
Yes! Thank you so much. I will be careful with the units.
And attempt to solution B, I was using an Impulse formula and I made a mistake.
When you try to solve for the Force. You must divide time from the force.
So It would be,
F(.19s)/.19s=(.25kg*27.78m/s)/.19s
F=36.55 N
 
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Likes   Reactions: RaulTheUCSCSlug
I think this is now correct.
 

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